| Usage: | |
| ```python | |
| import torch | |
| from transformers import AutoTokenizer, AutoModelForCausalLM | |
| question_template = "# Question\n\n{question}\n\n# Solution\n\n" | |
| model_name = "ScalableMath/llemma-7b-sft-prm800k-level-1to3-hf" | |
| model = AutoModelForCausalLM.from_pretrained(model_name, torch_dtype=torch.bfloat16, device_map="auto") | |
| tokenizer = AutoTokenizer.from_pretrained("EleutherAI/llemma_7b") | |
| question = "Convert the point $(0,3)$ in rectangular coordinates to polar coordinates. Enter your answer in the form $ | |
| (r,\\theta),$ where $r > 0$ and $0 \\le \\theta < 2 \\pi.$" | |
| question = question_template.format(question=question) | |
| input_tensor = torch.tensor([tokenizer.encode(question)]) | |
| outputs = model.generate(input_tensor.to(model.device), max_new_tokens=500) | |
| result = tokenizer.decode(outputs[0], skip_special_tokens=True) | |
| print(result) | |
| ``` | |
| Example Results: | |
| ``` | |
| # Question | |
| Convert the point $(0,3)$ in rectangular coordinates to polar coordinates. Enter your answer in the form $(r,\theta),$ where $r > 0$ and $0 \le \theta < 2 \pi.$ | |
| # Solution | |
| To convert from rectangular to polar coordinates, I need to use the formulas $r = \sqrt{x^2 + y^2}$ and $\theta = \tan^{-1}(y/x).$ | |
| In this case, $x = 0$ and $y = 3,$ so I can plug them into the formulas. | |
| For $r,$ I get $r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3.$ | |
| For $\theta,$ I get $\theta = \tan^{-1}(3/0).$ | |
| This is undefined, since the tangent function is not defined at $0.$ | |
| However, I can use the fact that the point $(0,3)$ lies on the positive $y$-axis, which has an angle of $\pi/2$ radians or $90^\circ.$ | |
| Therefore, I can choose any angle in the range $(0,\pi/2)$ as the value of $\theta.$ | |
| I will choose $\theta = \pi/2,$ since it is the simplest and most natural choice. | |
| Therefore, the polar coordinates of the point $(0,3)$ are $(3,\pi/2).$ | |
| # Answer | |
| (3,\pi/2) | |
| ``` |